Question

Question #f69d9

Answer 1

`y = 3x+12`

Explanation:

This problem is asking us to find a line such that the line intersects the graph of `y = x^3 + 4x^2 + 3x + 12` at the x and y intercepts of that graph (which would then also be the intercepts of that line).

To do this, all we have to do is find the `x` and `y` intercepts of the graph, and then draw the line passing through those two intercepts to complete the system.

Let's first solve for the `y` intercept by setting x equal to `0`.

`y = (0)^3 + 4(0)^2 + 3(0) + 12`

`y = 12`

So the y intercept is `12`, or `(0, 12)`. Now, we set y equal to `0` to find the `x` intercept:

`(0) = x^3 + 4x^2 + 3x + 12`

We can group the terms in groups of two, and then factor and redistribute to get the following:

`0 = (x^3 + 4x^2) + (3x + 12)`

`0 = x^2(x + 4) + 3(x+4)`

`0 = (x^2+3)(x+4)`

Since these two terms multiplied together gives us zero, this means that either `x^2+3` is equal to zero, or `x+4` is equal to zero. However, `x^2+3` must always be greater than 0, so our solution must be when:

`x+4 = 0`

`x = -4`

So our `x` intercept is `-4`, or `(-4, 0)`.

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Now, we have our two intercepts: `(-4,0)` and `(0, 12)`. The line connecting these two points will be our solution. Let's find the slope of this line:

`m = (y_2-y_1)/(x_2-x_1) = (12 - 0)/(0 - (-4)) = 12/4 = 3`

So our slope is 3, and our `y` intercept is 12. Using slope-intercept form, this means that the equation of the line is:

`y = 3x+12`

Final Answer